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Given a circle $P$ between two lines $\ell_1, \ell_2$, we want to find a tangent $AB$ to $P$ such that $A\in\ell_2, B\in\ell_1$ and the midpoint of $AB$ is the tangency point.

Is it possible to solve this problem through straightedge and compass?

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    $\begingroup$ Somehow I suspect there are two such tangents $\endgroup$ – Henry Jun 30 '17 at 13:51
  • $\begingroup$ one can use only compass and ruler? $\endgroup$ – serg_1 Jun 30 '17 at 14:24
  • $\begingroup$ hint. $PA=PB$, so "roughly" you can put compass in $P$ and and change points on $L_2$ draw arc to intersect $L_1$ before $AB$ will be tangent to the circle $\endgroup$ – serg_1 Jun 30 '17 at 14:39
  • $\begingroup$ Yes there are two tangents but how ? $\endgroup$ – MPO Jun 30 '17 at 16:04
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    $\begingroup$ Aren't there infinitely many tangents? Just imagine drawing tangents to this circle and extending it. These extensions will intersect both lines eventually? $\endgroup$ – David Jun 30 '17 at 17:00
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P1. Given a circle $\Gamma$ centered at $O$ and two points $A,B$ outside $\Gamma$,
find $C\in\Gamma$ such that the $OC$ line bisects the $\widehat{ACB}$ angle.

P2. Given a point $P$ in the first quadrant, find a point $A$ on the positive $y$-axis and a point $B$
on the positive $x$-axis such that $AB$ goes through $P$ and $AB$ is as short as possible.

Both these seemingly harmless problems (P1 has been studied by Leonardo Da Vinci) cannot be solved by straightedge and compass alone, because they boils down to finding the roots of a cubic polynomial (P2 is equivalent to the duplication of a cube).

Your problem is similar: we may consider a variable point $A$ on the $\ell_2$ line and draw the tangents from $A$ to the given circle, giving two tangency points $T_1,T_2$. Let $S_1$ be the symmetric of $A$ with respect to $T_1$ and let $S_2$ be the symmetric of $A$ with respect to $T_2$. Then $S_1$ lies on a curve $\gamma_1$, $S_2$ lies on a curve $\gamma_2$ and the problem is equivalent to finding the intersections of $\gamma_1,\gamma_2$ with $\ell_1$.
Since $\gamma_1$ and $\gamma_2$ are cubic curves, the given problem cannot be solved by straightedge and compass alone.

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  • $\begingroup$ so how can we solve it ? $\endgroup$ – MPO Jun 30 '17 at 20:04
  • $\begingroup$ @MPO: by finding the marked intersections, i.e. by finding the unique real root of two cubic polynomials. $\endgroup$ – Jack D'Aurizio Jun 30 '17 at 20:09
  • $\begingroup$ Perhaps your argument should say that $\gamma_1$ is a nondegenerate cubic. (Perhaps that's implicit in the language for you, but may not be for OP.) $\endgroup$ – John Hughes Aug 2 '17 at 12:14
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Let $r_1$ and $r_2$ straight lines so that: $$r_1: y =kx$$ and $$r_2: y =0.$$ Let $\Gamma$ a circle centered at $C_0=(x_0,y_0)$ with radius $r$.

Solve the following equation for $\theta$ and you will get the point T of tangency. $$ky_0\sin(\theta)+kr(\sin(\theta))^2=kx_0\cos(\theta)-2y_0\cos(\theta)+rk(\cos(\theta))^2-2r\sin(\theta)\cos(\theta)$$ $$T=(x_0+r\cos(\theta),y_0+r\sin(\theta)).$$ Draw tangent lines at T, and you will get the answer of the problem.

For $x_0=2$, $y_0=1$, $k=1$ and $ r=0.5$, you get $\theta=13.18$ degrees and $\theta=142.38$ degrees.

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Let $AB$ just be tangent with circle, and their midpoint $M$ goes into a hyperbola which is tangent to the circle. Now we aim to find the tangent point. Not sure if possible (we have two equations but only one unknown value, so can just remove some unnecessary part)

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